Consider The Function $f(x) = 2x^2 - 5x - 7$ On The Closed Interval \[-3, 4\].(a) Find Any Critical Values On The Interval $(-3, 4$\].No Decimal Entries Allowed. If There Is More Than One Critical Value, List Them Separated By

Introduction

In calculus, critical values are points on a function's graph where the derivative is zero or undefined. These points are significant because they can help us identify the function's maximum and minimum values. In this article, we will explore the critical values of the quadratic function $f(x) = 2x^2 - 5x - 7$ on the closed interval $[-3, 4]$.

The Function and Its Derivative

The given function is a quadratic function in the form $f(x) = ax^2 + bx + c$, where $a = 2$, $b = -5$, and $c = -7$. To find the critical values, we need to find the derivative of the function.

The derivative of $f(x)$ is given by:

$$f'(x) = \frac{d}{dx} (2x^2 - 5x - 7)$$

Using the power rule for differentiation, we get:

$$f'(x) = 4x - 5$$

Finding Critical Values

Critical values occur when the derivative is zero or undefined. In this case, the derivative is a linear function, so it is defined for all real numbers. Therefore, we need to find the values of $x$ that make the derivative equal to zero.

Setting the derivative equal to zero, we get:

$$4x - 5 = 0$$

Solving for $x$, we get:

$$x = \frac{5}{4}$$

So, the only critical value on the interval $(-3, 4)$ is $x = \frac{5}{4}$.

Discussion

In this article, we found the critical value of the quadratic function $f(x) = 2x^2 - 5x - 7$ on the closed interval $[-3, 4]$. The critical value is $x = \frac{5}{4}$, which is the only point on the interval where the derivative is zero.

Conclusion

In conclusion, the critical value of the quadratic function $f(x) = 2x^2 - 5x - 7$ on the closed interval $[-3, 4]$ is $x = \frac{5}{4}$. This value is significant because it can help us identify the function's maximum and minimum values on the given interval.

Additional Information

To find the maximum and minimum values of the function on the interval $[-3, 4]$, we need to evaluate the function at the critical value and at the endpoints of the interval.

Evaluating the function at the critical value, we get:

$$f\left(\frac{5}{4}\right) = 2\left(\frac{5}{4}\right)^2 - 5\left(\frac{5}{4}\right) - 7$$

Simplifying, we get:

$$f\left(\frac{5}{4}\right) = \frac{25}{8} - \frac{25}{4} - 7$$

$$f\left(\frac{5}{4}\right) = \frac{25}{8} - \frac{50}{8} - \frac{56}{8}$$

$$f\left(\frac{5}{4}\right) = -\frac{81}{8}$$

Evaluating the function at the endpoints of the interval, we get:

$$f(-3) = 2(-3)^2 - 5(-3) - 7$$

$$f(-3) = 18 + 15 - 7$$

$$f(-3) = 26$$

$$f(4) = 2(4)^2 - 5(4) - 7$$

$$f(4) = 32 - 20 - 7$$

$$f(4) = 5$$

So, the maximum value of the function on the interval $[-3, 4]$ is $f(-3) = 26$, and the minimum value is $f(4) = 5$.

Final Answer

$$$$$$$$$$

Introduction

In our previous article, we explored the critical values of the quadratic function $f(x) = 2x^2 - 5x - 7$ on the closed interval $[-3, 4]$. In this article, we will answer some frequently asked questions about critical values and provide additional information to help you better understand this concept.

Q&A

Q: What is a critical value?

A: A critical value is a point on a function's graph where the derivative is zero or undefined. These points are significant because they can help us identify the function's maximum and minimum values.

Q: How do I find critical values?

A: To find critical values, you need to find the derivative of the function and set it equal to zero. Then, solve for the value of $x$ that makes the derivative equal to zero.

Q: What if the derivative is undefined at a point?

A: If the derivative is undefined at a point, it means that the function has a vertical asymptote at that point. In this case, the point is not a critical value.

Q: Can there be more than one critical value?

A: Yes, there can be more than one critical value. In fact, a function can have multiple critical values, and each one can correspond to a maximum, minimum, or point of inflection.

Q: How do I determine whether a critical value corresponds to a maximum, minimum, or point of inflection?

A: To determine whether a critical value corresponds to a maximum, minimum, or point of inflection, you need to examine the behavior of the function around the critical value. You can do this by using the first derivative test or the second derivative test.

Q: What is the first derivative test?

A: The first derivative test is a method used to determine whether a critical value corresponds to a maximum, minimum, or point of inflection. To use the first derivative test, you need to evaluate the derivative of the function at a point on either side of the critical value. If the derivative changes from positive to negative, the critical value corresponds to a maximum. If the derivative changes from negative to positive, the critical value corresponds to a minimum. If the derivative does not change sign, the critical value corresponds to a point of inflection.

Q: What is the second derivative test?

A: The second derivative test is a method used to determine whether a critical value corresponds to a maximum, minimum, or point of inflection. To use the second derivative test, you need to evaluate the second derivative of the function at the critical value. If the second derivative is positive, the critical value corresponds to a minimum. If the second derivative is negative, the critical value corresponds to a maximum. If the second derivative is zero, the critical value corresponds to a point of inflection.

Q: Can I use the second derivative test if the second derivative is undefined?

A: No, you cannot use the second derivative test if the second derivative is undefined. In this case, you need to use the first derivative test or another method to determine whether the critical value corresponds to a maximum, minimum, or point of inflection.

Q: What if I have a function with multiple critical values?

A: If you have a function with multiple critical values, you need to examine the behavior of the function around each critical value to determine whether it corresponds to a maximum, minimum, or point of inflection.

Conclusion

In this article, we answered some frequently asked questions about critical values and provided additional information to help you better understand this concept. We hope that this article has been helpful in clarifying any confusion you may have had about critical values.

Additional Resources

If you are interested in learning more about critical values, we recommend the following resources:

• Calculus textbook
• Online calculus course
• Mathematics website

Final Answer

The critical value of the quadratic function $f(x) = 2x^2 - 5x - 7$ on the closed interval $[-3, 4]$ is $x = \frac{5}{4}$. The maximum value of the function on the interval is $f(-3) = 26$, and the minimum value is $f(4) = 5$.